Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.
a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.
b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

Answer 1

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: $sin \theta=\frac{m\lambda}{\alpha}$

Given $m=3$,$\lambda=6.33\times 10^-^7$ and $\alpha=0.00015$:

#Substitute our variable values in the minima equation to obtain $\theta$:

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

#draw a triangle to find the relationship between $\theta, y$ and $L$.

$tan(\theta)=y/L$               #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b.  To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$, where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$=4.5cm

Then from the equation subtract $y_3$ from $y$:

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm